One hundred grams of water is saturated with NH CI at
50°C. According to Table G. if the temperature is lowered
to 10° c, what is the total amount of NH4Cl that will
precipitate?

Explanation:

The table G is the graph of the solubility curves for several solutes which is attached.

The second picture identifies the solubilities for the  NH₄Cl at 50ºC and 10ºC.

The solubility of NH₄Cl at 50ºC is about 52 g/ 100 g of water.

The solubility of NH₄Cl at 10ºC is about 34 g / 100 g of water.

Then, at 50ºC 100 g of water saturated with NH₄Cl contains about 52 g of NH₄Cl and 100 g of water saturated with NH₄Cl contains 34 g of NH₄Cl.

The difference, 52g - 34 g of NH₄Cl shall precipitate:

          52 g - 34 g = 18g ← answer

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

the dissolution of polymer can be thermodynamically described in 2 steps.